A property of the derivatives of a function Suppose that $f,g_1,g_2,\dots$ are functions from $\mathbb{R}$ to $\mathbb{R}$ such that $f'=f\,g_1$ and $g'_j=g_j^2-g_j g_{j+1}$. Here and in what follows, $j$ is any natural number. Then, by induction, $f^{(j)}=f\, P_j(g_1,\dots,g_j)$, where 
$$P_j(u_1,\dots,u_j)=\sum_{k_1=0}^j\cdots\sum_{k_j=0}^j c_{k_1,\dots,k_j}
u_1^{k_1}\cdots u_j^{k_j}
$$
is a polynomial in $\mathbb{R}[u_1,\dots,u_j]$ of degree $j$. 
The problem is to show that 
$$\sum_{k_1=0}^j\cdots\sum_{k_j=0}^j |c_{k_1,\dots,k_j}|\, 
1^{k_1}2^{k_2}\cdots j^{k_j}=2^{j - 1} j! 
$$
for all natural $j$. 
This equality has been verified for $j=1,\dots,10$.
 A: We have, writing for short $P_n=P_n(g_1,\dots,g_{n })$,
$$\big(f^{(n)}\big)' =\big(fP_{n }\big)'=f' P_{n }+\sum_{i=1}^n f\partial_iP_ng_i'=fg_1 P_{n }+\sum_{i=1}^n f\partial_iP_n(g_i^2-g_ig_{i+1})$$
Comparing with $f^{(n+1)}=fP_{n+1}$ we obtain a linear recursion for the sequence of polynomials $P_n=P_n(x_1,\dots,x_{n })$
\begin{cases} P_1=x_1  \\ P_{n+1}=x_1P_n  +\sum_{i=1}^n (x_i - x_{i+1}) x_i\partial_i P_n, \end{cases} 
whence it is clear that $P_n$ is a homogeneous polynomial of degree $n$, $P_ n:=\sum_{|\alpha|=n} c(\alpha)x^\alpha$, the sum being extended over all multi-indices $\alpha:=(\alpha_1,\alpha_2,\dots )$ of weight 
$|\alpha|:=\alpha_1+\alpha_2+\dots=n.$
Equating the coefficients of the monomial $x^\alpha:=x_1^{\alpha_1}\dots x_{n+1}^{\alpha_{n+1}} $ of degree $n+1:=|\alpha|$ in the above recursive formulas  we obtain:
$$c(\alpha)= \alpha_1 c(\alpha-\delta_1)- \sum_{i=2}^{n} (\alpha_{i-1}-\alpha_{i }+1)c(\alpha-\delta_i) \; - \alpha_{n}c(\alpha-\delta_{n+1}),$$
where $\delta_i:=(\delta_{i1},\delta_{i2},\delta_{i3}\dots)$ and $\delta_{ij}$ is the usual Kronecker symbol.
From this formula for $c(\alpha)$ it follows easily by induction that
(i) If $c(\alpha)\neq0$ then $\alpha_1\ge\alpha_2\ge \dots  $;
(ii) $\operatorname{sgn}c(\alpha)=(-1)^{|\alpha|-\alpha_1}$.
As a consequence, the  sum we are interested in is  $$ \sum_{|\alpha|=n} \big|c(\alpha)\big|1^{\alpha_1}2^{\alpha_2}\dots n^{\alpha_n}=P_n( 1, - 2,- 3,\dots,- n).$$
To evaluate it, consider $\sigma_n(t):=P_n( t, - 2,- 3,\dots,- n).$ Then, by the recurrence relation of the polynomials $P_n$ 
and by the Euler formula for homogeneous polynomials we find
\begin{cases} \sigma_1=t \\ \sigma_{n+1} =(t+n)\sigma_n +  (t+t^2)\sigma_n' , \end{cases} 
Now it is easy to check that
$$\sigma_n(t)=n!t(t+1)^{n-1},$$
whence $\sigma_n(1)=n!2^{n-1}.$ 
$$*$$
rmk. Also note that the polynomials $Q_n:=  \sum_{|\alpha|=n} \big|c(\alpha)\big|x^\alpha$ are simply $P_n(x_1, -x_2,-x_3,\dots,-x_n),$ whence one finds a linear recurrence relation for them 
\begin{cases} Q_1=x_1  \\ Q_{n+1}=x_1Q_n+2x_1^2\partial_1Q_n  +\sum_{i=1}^n (x_{i+1} - x_{i}) x_i\partial_i Q_n. \end{cases} 
